Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
ECG rhythm strip interpretation
Paper (25 mm/s, 10 mm/mV)
1 small box = 0.04 s, 1 large box = 0.20 s, 10 mm = 1 mV
Heart rate, regular rhythm
300 ÷ large boxes = 1500 ÷ small boxes = 60 ÷ RR (s)
Heart rate, irregular rhythm
QRS complexes in 6 s × 10
Normal PR interval
0.12–0.20 s
Normal QRS duration
< 0.12 s
QTc (Bazett)
QT/√RR
QTc (Fridericia)
QT/∛RR
QTc (Framingham)
QT + 154·(1 − RR)
Sinus rates
bradycardia < 60, normal 60–100, tachycardia > 100 /min
Escape rates
junctional 40–60, ventricular 20–40 /min
Fast rhythms
SVT 150–250 /min (narrow), VT > 100 /min (wide)
Method
Rate: for a regular rhythm divide 300 by the large boxes or 1500 by the small boxes between two R waves; for an irregular rhythm count the QRS complexes in 6 seconds and multiply by 10.
Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
P waves: is there an upright P wave in lead II before every QRS, all alike? Look for fibrillatory or sawtooth waves, inverted P waves and P waves without a QRS.
PR interval: from the start of P to the start of the QRS, normally 0.12–0.20 s and constant. Watch whether it lengthens, stays fixed before a dropped beat, or varies at random.
QRS duration: below 0.12 s is narrow; 0.12 s or more is wide. Then name the rhythm, and for the QT use QTc = QT/√RR (Bazett) with times in seconds.
Evidence-based practice statistics
Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time
Method
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.
2Practice problems
1.Human anatomyMedium
The sympathetic nervous system leaves the central nervous system from which spinal segments?
ACervical only
BThoracic and upper lumbar (T1–L2)
CSacral only (S2–S4)
DCranial and sacral
2.Human anatomyHard
A large central disc prolapse at L4–L5 causes saddle anaesthesia, urinary retention and bilateral leg weakness. Which structure is compressed?
AFemoral nerve
BThoracic spinal cord
CSciatic nerve in the buttock
DCauda equina
3.Human physiologyMedium
How is the ejection fraction calculated?
AStroke volume ÷ end-diastolic volume
BEnd-systolic volume ÷ stroke volume
CCardiac output ÷ heart rate
DEnd-diastolic volume ÷ end-systolic volume
4.Human physiologyHard
After several days of persistent vomiting of stomach contents, which acid–base disturbance is most likely?
AMetabolic acidosis
BRespiratory acidosis
CMetabolic alkalosis
DRespiratory alkalosis
5.MicrobiologyMedium
Transfer of bacterial genes from one bacterium to another by a virus is called:
ATranscription
BConjugation
CTransformation
DTransduction
6.MicrobiologyHard
A child develops bloody diarrhoea after eating an undercooked burger, then a week later has kidney failure, anaemia and a low platelet count. Which organism is most likely?
AClostridium perfringens
BShiga toxin-producing E. coli such as O157:H7
CStaphylococcus aureus
DRotavirus
7.PharmacologyMedium
How should a hyoscine (scopolamine) patch for motion sickness be used?
AApply it behind the ear several hours before travel
BSwallow it if no water is available
CApply it to the chest only after the vomiting has started
DCut it in half for children without advice
8.PharmacologyHard
A breastfeeding mother needs a mild analgesic. Which drug is generally considered compatible with breastfeeding?
AAspirin
BParacetamol (acetaminophen)
CCodeine (as co-codamol)
DPethidine (meperidine)
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 73 mmHg (9.7 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic alkalosis with HCO₃⁻ 34 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 34 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Fridericia’s formula QTc = QT/∛RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
12.ECG rhythm strip interpretationHard
A monitored patient is found unresponsive and not breathing normally, and no carotid pulse can be felt. The monitor shows this organised rhythm. What is this situation called? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1asystole
2ventricular fibrillation
3sinus bradycardia
4pulseless electrical activity (PEA)
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 1000 patients received a new treatment and 400 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the absolute risk reduction ARR as a percentage. Round to one decimal place.
treatment: 195/1000control: 156/400
ARR = %
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 20% for a disease. A test with sensitivity 82% and specificity 86% comes back positive. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 20%Se = 82%Sp = 86%positive
p′ = %
3Solutions
1.Human anatomy
Answer: B · Thoracic and upper lumbar (T1–L2)
Why this is the answer
Sympathetic preganglionic neurons lie in the lateral horn from T1 to L2. The parasympathetic system has a craniosacral outflow.
Answer
B · Thoracic and upper lumbar (T1–L2)
2.Human anatomy
Answer: D · Cauda equina
Why this is the answer
Below L1–L2 the canal holds the cauda equina. Central compression of these lumbar and sacral roots causes cauda equina syndrome, a surgical emergency.
Answer
D · Cauda equina
3.Human physiology
Answer: A · Stroke volume ÷ end-diastolic volume
Why this is the answer
Ejection fraction is the proportion of blood in the full ventricle that is ejected per beat. A normal value is about 55–70%.
Answer
A · Stroke volume ÷ end-diastolic volume
4.Human physiology
Answer: C · Metabolic alkalosis
Why this is the answer
Vomiting removes HCl, and the stomach's acid secretion leaves bicarbonate in the blood. Volume loss and low potassium help keep the alkalosis going.
Answer
C · Metabolic alkalosis
5.Microbiology
Answer: D · Transduction
Why this is the answer
In transduction a bacteriophage accidentally packages bacterial DNA and delivers it to the next cell it infects. Transformation is uptake of free DNA and conjugation uses direct contact.
Answer
D · Transduction
6.Microbiology
Answer: B · Shiga toxin-producing E. coli such as O157:H7
Why this is the answer
Shiga toxin-producing E. coli damages the bowel and the small blood vessels of the kidney, causing haemolytic uraemic syndrome. Cattle are a reservoir, so undercooked minced beef is a classic source.
Answer
B · Shiga toxin-producing E. coli such as O157:H7
7.Pharmacology
Answer: A · Apply it behind the ear several hours before travel
Why this is the answer
The patch is applied behind the ear several hours before travel, and the hands are washed afterwards. Touching the eye with drug on the fingers can dilate the pupil and blur vision.
Answer
A · Apply it behind the ear several hours before travel
8.Pharmacology
Answer: B · Paracetamol (acetaminophen)
Why this is the answer
Paracetamol passes into milk in small amounts and is widely regarded as safe. Codeine can cause dangerous opioid levels in babies of ultra-rapid metaboliser mothers, aspirin carries a Reye's risk and pethidine's metabolite accumulates in infants.
Answer
B · Paracetamol (acetaminophen)
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 27.3 mmol/L
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Change in PaCO₂
|73 − 40| = 33 mmHg → 3.3 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·3.3 = 27.3 mmol/L
Answer
HCO₃⁻ = 27.3 mmol/L
10.Arterial blood gas interpretation
Answer: PaCO₂ = 44.8 mmHg
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Expected PaCO₂
0.7·34 + 21 = 44.8 mmHg
Answer
PaCO₂ = 44.8 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: QTc = 0.52 s
QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.
Measure
QT = 13 × 0.04 = 0.52 s
RR = 25 × 0.04 = 1.00 s
Fridericia’s formula
QTc = 0.52/∛1.00 = 0.52/1 = 0.52 s
Answer
QTc = 0.52 s
Above 0.50 s: markedly prolonged. (520 ms)
12.ECG rhythm strip interpretation
Answer: pulseless electrical activity (PEA)
Pulseless electrical activity (PEA): PEA cannot be seen on a strip. The monitor shows organised electrical activity (a rhythm other than VF, VT or asystole) but the patient has no palpable pulse, so it is recognised by checking the patient. In adult resuscitation guidelines PEA and asystole are the non-shockable cardiac arrest rhythms: high-quality CPR, epinephrine as soon as possible and a search for reversible causes (the Hs and Ts, such as hypovolaemia, hypoxia, tension pneumothorax and cardiac tamponade). VF and pulseless VT are the shockable rhythms.
What the strip shows
1. Rate: R–R = 36 small boxes, 1500 ÷ 36 ≈ 42 per minute.
2. Rhythm: regular, every R–R is 1.44 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.16 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
sinus bradycardia
What the patient shows
no pulse despite organised electrical activity: pulseless electrical activity (PEA), a non-shockable cardiac arrest rhythm
Answer
pulseless electrical activity (PEA)
Answer
pulseless electrical activity (PEA)
PEA is recognised at the bedside, never from the strip alone. This is exam practice, not clinical guidance.
13.Evidence-based practice statistics
Answer: ARR = 19.5 %
Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.
Event rates
EER = 195/1000 = 0.195
CER = 156/400 = 0.39
Absolute risk reduction
ARR = 0.39 − 0.195 = 0.195 = 19.5%
Answer
ARR = 19.5%
14.Evidence-based practice statistics
Answer: p′ = 59.4 %
Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.